Problem: The lifespans of seals in a particular zoo are normally distributed. The average seal lives $12.9$ years; the standard deviation is $3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a seal living longer than $9.9$ years.
Solution: $12.9$ $9.9$ $15.9$ $6.9$ $18.9$ $3.9$ $21.9$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $12.9$ years. We know the standard deviation is $3$ years, so one standard deviation below the mean is $9.9$ years and one standard deviation above the mean is $15.9$ years. Two standard deviations below the mean is $6.9$ years and two standard deviations above the mean is $18.9$ years. Three standard deviations below the mean is $3.9$ years and three standard deviations above the mean is $21.9$ years. We are interested in the probability of a seal living longer than $9.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the seals will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the seals will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $9.9$ years and the other half $({16\%})$ will live longer than $15.9$ years. The probability of a particular seal living longer than $9.9$ years is ${68\%} + {16\%}$, or $84\%$.